Theorem unineq 3506

Description: Infer equality from equalities of union and intersection. Exercise 20 of [Enderton] p. 32 and its converse. (Contributed by NM, 10-Aug-2004.)

Assertion

Ref	Expression
unineq	⊢ (((A ∪ C) = (B ∪ C) ∧ (A ∩ C) = (B ∩ C)) ↔ A = B)

Proof of Theorem unineq

Dummy variable x is distinct from all other variables.

Step	Hyp	Ref	Expression
1		eleq2 2414	. . . . . . 7 ⊢ ((A ∩ C) = (B ∩ C) → (x ∈ (A ∩ C) ↔ x ∈ (B ∩ C)))
2		elin 3220	. . . . . . 7 ⊢ (x ∈ (A ∩ C) ↔ (x ∈ A ∧ x ∈ C))
3		elin 3220	. . . . . . 7 ⊢ (x ∈ (B ∩ C) ↔ (x ∈ B ∧ x ∈ C))
4	1, 2, 3	3bitr3g 278	. . . . . 6 ⊢ ((A ∩ C) = (B ∩ C) → ((x ∈ A ∧ x ∈ C) ↔ (x ∈ B ∧ x ∈ C)))
5		iba 489	. . . . . . 7 ⊢ (x ∈ C → (x ∈ A ↔ (x ∈ A ∧ x ∈ C)))
6		iba 489	. . . . . . 7 ⊢ (x ∈ C → (x ∈ B ↔ (x ∈ B ∧ x ∈ C)))
7	5, 6	bibi12d 312	. . . . . 6 ⊢ (x ∈ C → ((x ∈ A ↔ x ∈ B) ↔ ((x ∈ A ∧ x ∈ C) ↔ (x ∈ B ∧ x ∈ C))))
8	4, 7	syl5ibr 212	. . . . 5 ⊢ (x ∈ C → ((A ∩ C) = (B ∩ C) → (x ∈ A ↔ x ∈ B)))
9	8	adantld 453	. . . 4 ⊢ (x ∈ C → (((A ∪ C) = (B ∪ C) ∧ (A ∩ C) = (B ∩ C)) → (x ∈ A ↔ x ∈ B)))
10		uncom 3409	. . . . . . . . 9 ⊢ (A ∪ C) = (C ∪ A)
11		uncom 3409	. . . . . . . . 9 ⊢ (B ∪ C) = (C ∪ B)
12	10, 11	eqeq12i 2366	. . . . . . . 8 ⊢ ((A ∪ C) = (B ∪ C) ↔ (C ∪ A) = (C ∪ B))
13		eleq2 2414	. . . . . . . 8 ⊢ ((C ∪ A) = (C ∪ B) → (x ∈ (C ∪ A) ↔ x ∈ (C ∪ B)))
14	12, 13	sylbi 187	. . . . . . 7 ⊢ ((A ∪ C) = (B ∪ C) → (x ∈ (C ∪ A) ↔ x ∈ (C ∪ B)))
15		elun 3221	. . . . . . 7 ⊢ (x ∈ (C ∪ A) ↔ (x ∈ C ∨ x ∈ A))
16		elun 3221	. . . . . . 7 ⊢ (x ∈ (C ∪ B) ↔ (x ∈ C ∨ x ∈ B))
17	14, 15, 16	3bitr3g 278	. . . . . 6 ⊢ ((A ∪ C) = (B ∪ C) → ((x ∈ C ∨ x ∈ A) ↔ (x ∈ C ∨ x ∈ B)))
18		biorf 394	. . . . . . 7 ⊢ (¬ x ∈ C → (x ∈ A ↔ (x ∈ C ∨ x ∈ A)))
19		biorf 394	. . . . . . 7 ⊢ (¬ x ∈ C → (x ∈ B ↔ (x ∈ C ∨ x ∈ B)))
20	18, 19	bibi12d 312	. . . . . 6 ⊢ (¬ x ∈ C → ((x ∈ A ↔ x ∈ B) ↔ ((x ∈ C ∨ x ∈ A) ↔ (x ∈ C ∨ x ∈ B))))
21	17, 20	syl5ibr 212	. . . . 5 ⊢ (¬ x ∈ C → ((A ∪ C) = (B ∪ C) → (x ∈ A ↔ x ∈ B)))
22	21	adantrd 454	. . . 4 ⊢ (¬ x ∈ C → (((A ∪ C) = (B ∪ C) ∧ (A ∩ C) = (B ∩ C)) → (x ∈ A ↔ x ∈ B)))
23	9, 22	pm2.61i 156	. . 3 ⊢ (((A ∪ C) = (B ∪ C) ∧ (A ∩ C) = (B ∩ C)) → (x ∈ A ↔ x ∈ B))
24	23	eqrdv 2351	. 2 ⊢ (((A ∪ C) = (B ∪ C) ∧ (A ∩ C) = (B ∩ C)) → A = B)
25		uneq1 3412	. . 3 ⊢ (A = B → (A ∪ C) = (B ∪ C))
26		ineq1 3451	. . 3 ⊢ (A = B → (A ∩ C) = (B ∩ C))
27	25, 26	jca 518	. 2 ⊢ (A = B → ((A ∪ C) = (B ∪ C) ∧ (A ∩ C) = (B ∩ C)))
28	24, 27	impbii 180	1 ⊢ (((A ∪ C) = (B ∪ C) ∧ (A ∩ C) = (B ∩ C)) ↔ A = B)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 176   ∨ wo 357   ∧ wa 358   = wceq 1642   ∈ wcel 1710   ∪ cun 3208   ∩ cin 3209

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-un 3215

This theorem is used by:  phiall  4619